Calculate the pressure that a certain amount of water applies

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Discussion Overview

The discussion revolves around calculating the pressure that water applies on the wall of a dam, exploring the relevant equations and methods for determining pressure distribution. Participants examine the implications of atmospheric pressure and the geometric approach to calculating net force on the dam.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the equation Pressure = P(atmosphere) + (density(water) * height * gravity) applies only to the bottom of the system.
  • Another participant confirms the equation is correct but clarifies that "height" refers to depth below the water surface.
  • Some participants suggest integrating the pressure distribution along the wall or using geometric methods to find the area of a triangle.
  • There is a discussion about the pressure at the bottom being 392000 Pa and the pressure at the top being zero, leading to a linear variation of pressure.
  • One participant expresses confusion about whether atmospheric pressure should be included in their calculations, noting they calculated an average pressure and divided it by two.
  • Another participant explains that atmospheric pressure can be excluded since it acts on both sides of the dam, thus canceling out.
  • Some participants emphasize the importance of understanding the derivation of the pressure calculations rather than just applying formulas blindly.
  • There is a mention of the center of pressure concept, indicating that water exerts a pressure distribution rather than a single pressure value.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of including atmospheric pressure in calculations, with some arguing it can be omitted while others assert it must be included for net force calculations. The discussion remains unresolved regarding the best approach to calculating pressure on the dam.

Contextual Notes

Participants have not reached a consensus on the method for calculating the net force on the dam, and there are varying interpretations of how to apply atmospheric pressure in the context of the problem.

MartaBraga
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How do I calculate the pressure that a certain amount of water applies on the wall of a dem(example)?

Is the equation Pressure = P(atmosphere) + (density(water).height.gravity) correct? Or does that apply only for the pressure aplied on the bottom of the system?

(I'm sorry about the bad english)
 
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That equation is correct. Realize that "height" is really depth below the water surface.
 
.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.
 
Thats the correct formula for the pressure at that depth.
(the force sideways is the same as the force downwards because it is a fluid, so pushes in all directions)

Cyrus's reply is about calculating the TOTAL force on the wall.
At the bottom is the pressure you calculated, at the surface there is obviously no pressure and half way down is half the pressure.
 
Cyrus said:
.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.


How do I do that, I mean, calculate it through the area of the triangle? (I rather do anything than integrate!)
 
The pressuer at the bottom is: [tex]P=\rho * g*h[/tex]. The pressure at the top is zero. It varies linearly. I'll leave it to you to show us the rest of the derivation and logic. :smile:
 
Ok, so is it that simple?
I calculated the pressure through that formula and then divided it by 2 and multiplied it by the area... and the result is correct... But I thought I had to use the atmospheric pressure in the formula...

Or is this pure luck?
 
Can you show your steps. It would be more insightful if you knew why your answer is correct. Right now, you are just blindly plugging away and hoping for the best to come out.
 
Ok, the problem is to calculate the pressure the water applies on a dem, with 40m of heigth and 150m of lenght.
I calculated the pressure = 1000 (the density) x 40 (the height) x 9,8 (the aceleration of gravity. The result is 392000, which I divided by 2 = 196000. I multiplied this value by the area of the dem in contact with the water, 6000m2. The result is 1,176x10^9.
This was the result I was looking for.
 
  • #10
Ok, first of all the water does not exert a 'pressure' on the dam. It has a pressure distribution. This causes a net force acting at the center of pressure. (Think center of gravity, same concept different name).

Yes that exactly right. Do you know what you are doing though?
 
  • #11
That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, let's find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)
 
  • #12
"But I thought I had to use the atmospheric pressure in the formula..."

You can leave out the atmospheric pressure because it is applied to the downstream side of the dam also - hence it cancels out. If there were no water in the lake at all, just the air on both sides, there would be no net force on the dam.
 
  • #13
If the problem is to calculate the net force that the water exerts on the dam, then you must include atmospheric pressure. (But as gmax137 states, for some purposes you can forget atmospheric pressure when it acts on both sides of the dam.)
 
  • #14
MartaBraga said:
That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, let's find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)

Hint: The area of a triangle is 1/2*(base)*(height)!
 

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